In solving riemann sums, what is the difference between using right endpoints, left endpoints, and midpoints?

Yes.

When you partition an interval [a, b] into [a_1, a_2], [a_2, a_3], …, [a_n, a_(n+1)], the left endpoints are a_2, a_3, …, a_(n+1), hence the riemann sum for right endpoints will be:

f(a_2)*∆x_1 + f(a_3)*∆x_2 + … + f(a_(n+1))*∆x_n = (i=1 to n)∑ f(x_(i+1))*∆x_i

However, the left endpoints would be a_1, b_2, …, a_n, hence the riemann sum for the left endpoints would be:

f(a_1)*∆x_1 + f(a_2)*∆x_2 + … + f(a_n)*∆x_n = (i=1 to n)∑ f(x_i)*∆x_i

For the midpoint, take the average of the endpoints:

f((a_1 + a_2)/2)*∆x_1 + f((a_2 + a_3)/2)*∆x_2 + … + f((a_n + a_(n+1))/2)*∆x_n = (i=1 to n)∑ f((a_i + a_(i+1))/2)*∆x_i

But in the limit as n approaches infinity, they all come out to be the same (the area under the curve).

I hope this helps!

Let's say you use Riemann sums to approximate area under curve of function f(x) on interval [0, 10] using 5 subintervals of width (10-0)/5 = 2

The 5 subintervals are: [0,2], [2,4], [4,6], [6,8], [8,10]

We approximate area under curve by finding an approximation for each rectangle and adding these areas.

Right endpoints of each subintervals are at x = 2, 4, 6, 8, 10 respectively

So Riemann sum using right endpoints:

A ≈ 2 * f(2) + 2 * f(4) + 2 * f(6) + 2 * f(8) + 2 * f(10)

A ≈ 2 [f(2) + f(4) + f(6) + f(8) + f(10)]

Left endpoints of each subintervals are at x = 0, 2, 4, 6, 8 respectively

So Riemann sum using left endpoints:

A ≈ 2 * f(0) + 2 * f(2) + 2 * f(4) + 2 * f(6) + 2 * f(8)

A ≈ 2 [f(0) + f(2) + f(4) + f(6) + f(8)]

Midpoints of each subintervals are at x = 1, 3, 5, 7, 9 respectively

So Riemann sum using midpoints:

A ≈ 2 * f(1) + 2 * f(3) + 2 * f(5) + 2 * f(7) + 2 * f(9)

A ≈ 2 [f(1) + f(3) + f(5) + f(7) + f(9)]

You can also use trapezoidal sum, which is average of right endpoints and left endpoints

NOTE:

When the function f(x) is increasing, then Riemann sum using left endpoints gives approximation that is smaller than the actual area, and Riemann sum using right endpoints gives approximation that is larger than actual area.

When the function f(x) is decreasing, then Riemann sum using left endpoints gives approximation that is larger than the actual area, and Riemann sum using right endpoints gives approximation that is smaller than actual area.

When function f(x) is concave down, midpoint sum is larger than actual area and trapezoidal sum is smaller than actual area.

When function f(x) is concave up, midpoint sum is smaller than actual area and trapezoidal sum is larger than actual area.

Riemann sum using midpoints is more accurate than trapezoidal sum, which is more accurate than Riemann sum using left or right endpoints

Right Endpoint Riemann Sum

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RE:

In solving riemann sums, what is the difference between using right endpoints, left endpoints, and midpoints?

Is there a different equation for them? I'm really confused.

The left endpoint, right endpoint and midpoint will generally lead to different partial sums. However, if the function is integrable the sums will always converge to the same number. In general, one can select any points within each subinterval but the end points or mid points make the summation easier to work with.

Yes, there's a difference. Let f(x) = x+1 from x = 0 to 2. Use 2 Riemann sums (i.e., two rectangles of width = 1).

Left-hand sum = f(0) + f(1)

Right-hand sum = f(1) + f(2)

Midpoint = f(1/2) + f(3/2)

The left end points will give you an approximation which is a bit smaller then the actual value, the right end points an approximation which is a bit higher. So the closer one is the midpoints.

The approximations may be different using these choices.

However, the limits as the width of the rectangles become smaller and smaller will approach the same limit which is the definite integral of the function.